A \(20 \mathrm{~mm} \times 20 \mathrm{~mm}\) silicon chip is mounted such that the edges are flush in a substrate. The substrate provides an unheated starting length of \(20 \mathrm{~mm}\) that acts as turbulator. Airflow at \(25^{\circ} \mathrm{C}(1 \mathrm{~atm})\) with a velocity of \(25 \mathrm{~m} / \mathrm{s}\) is used to cool the upper surface of the chip. If the maximum surface temperature of the chip cannot exceed \(75^{\circ} \mathrm{C}\), determine the maximum allowable power dissipation on the chip surface.

We're given the temperature (\(T_\infty = 25^{\circ} \mathrm{C}\)) and pressure (\(P_\infty = 1 \mathrm{~atm}\)) of the air in the system. At this pressure and temperature, we can look up the air's density (\(\rho\)), dynamic viscosity (\(\mu\)), thermal conductivity (\(k\)), and specific heat capacity (\(c_p\)) in a standard reference. The following properties can be found: \(\rho = 1.184 \mathrm{~kg/m^{3}}\) \(\mu = 1.846 \times 10^{-5} \mathrm{~Pa.s}\) \(k = 2.602 \times 10^{-2} \mathrm{~W/m.K}\) \(c_p = 1006 \mathrm{~J/kg.K}\)

The Reynolds number is calculated using the following formula: \(Re = \frac{\rho V L}{\mu}\) Where \(V = 25 \mathrm{~m/s}\) is the velocity of the air, \(L = 20 \mathrm{~mm}\) is the length of the chip, and \(\rho\) and \(\mu\) are the air properties we found in step 1. \(Re = \frac{(1.184 \mathrm{~kg/m^{3}}) \times (25 \mathrm{~m/s}) \times (20 \times 10^{-3} \mathrm{~m})}{(1.846 \times 10^{-5} \mathrm{~Pa.s})} = 32046\)

For laminar flow over a flat plate, with a Reynolds number between 5000 and \(10^5\), we can use the following Nusselt number relation: \(Nu = 0.664Re^{0.5}Pr^{1/3}\) Where \(Nu\) is the Nusselt number, \(Re\) is the Reynolds number calculated in step 2, and \(Pr\) is the Prandtl number. The Prandtl number can be calculated using the following formula: \(Pr = \frac{c_p \mu}{k}\) \(Pr = \frac{(1006 \mathrm{~J/kg.K}) \times (1.846 \times 10^{-5} \mathrm{~Pa.s})}{(2.602 \times 10^{-2} \mathrm{~W/m.K})} = 0.707\) Inserting the values of \(Re\) and \(Pr\) into the Nusselt number relation: \(Nu = 0.664(32046)^{0.5}(0.707)^{1/3} = 217.8\) Now we can calculate the heat transfer coefficient (h) using the following formula: \(h = \frac{Nu \times k}{L}\) \(h = \frac{(217.8) \times (2.602 \times 10^{-2} \mathrm{~W/m.K})}{(20 \times 10^{-3} \mathrm{~m})} = 283.6 \mathrm{~W/m^2.K}\)

We can calculate the maximum allowable heat flux using the following relation: \(q'' = h(T_{max} - T_\infty)\) Where \(q''\) is the heat flux, \(h\) is the heat transfer coefficient we calculated in step 3, \(T_{max}\) is the maximum surface temperature (\(75^{\circ} \mathrm{C}\)), and \(T_\infty\) is the temperature of the air (\(25^{\circ} \mathrm{C}\)). \(q'' = (283.6 \mathrm{~W/m^2.K})(75-25) = 14180 \mathrm{~W/m^2}\)

The maximum allowable power dissipation can be calculated using the following formula: \(P_{max} = q'' \times A\) Where \(P_{max}\) is the maximum allowable power dissipation, \(q''\) is the heat flux calculated in step 4, and \(A\) is the surface area of the chip. Since the chip is \(20 \mathrm{~mm} \times 20 \mathrm{~mm}\), the surface area is: \(A = (20 \times 10^{-3} \mathrm{~m}) \times (20 \times 10^{-3} \mathrm{~m}) = 4 \times 10^{-4} \mathrm{~m^2}\) Now, we can calculate the maximum allowable power dissipation: \(P_{max} = (14180 \mathrm{~W/m^2}) \times (4 \times 10^{-4} \mathrm{~m^2}) = 5.67 \mathrm{~W}\) Therefore, the maximum allowable power dissipation on the chip surface is 5.67 W.